special relativity - How come $frac{partial(partial_{beta}A_{gamma})}{partial(partial_{mu}A_{nu})} = g_{betamu}g_{gammanu}$?

For context, this equation is used in the following (from Schwartz's QFT 3.44)

$$\partial_{\mu} \frac{\partial(\partial_{\beta}A_{\gamma})^2}{\partial(\partial_{\mu}A_{\nu})} = \partial_{\mu}\left[2(\partial_{\alpha}A_{\alpha})\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} g_{\beta\gamma}\right] = \partial_{\mu}[2(\partial_{\alpha}A_{\alpha})g_{\beta\mu}g_{\gamma\nu}g_{\beta\gamma}]=2\partial_{\nu}(\partial_{\alpha}A_{\alpha}).\tag{3.44}$$

But doesn't $$\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = g_{\beta\mu}g_{\gamma\nu}$$ imply, for instance, that $$\frac{\partial(\partial_0 A_3)}{\partial(\partial_0 A_3)} = g_{00}g_{33} = -1$$ instead of $=1$? Shouldn't we instead have $\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = \delta_{\beta\mu}\delta_{\gamma\nu}$?

Answer

Schwartz doesn't keep track of index placement. Assuming the indices on the left-hand side are placed correctly, fixing the right-hand side gives $$\frac{\partial(\partial_{\beta}A_{\gamma})}{\partial(\partial_{\mu}A_{\nu})} = g_{\beta}^{\ \ \mu}g_{\gamma}^{\ \ \nu}.$$ That is, you get the metric with mixed indices, and its diagonal components are all $+1$, not a mix of $+1$ and $-1$. (You can calculate that explicitly; another way to see this is that you get to the mixed metric by raising an index on the metric by contracting with the inverse metric. In components, multiplying a matrix with its inverse gives the identity matrix.) This fixes the sign error.

You might ask why Schwartz didn't just use $\delta$ instead of $g$. That's just kind of personal preference -- the $\delta$ is usually only considered with mixed indices, while $g$ is used with indices in all positions. So Schwartz's expression generalizes to arbitrary index placement nicely.

The general moral is that it's fine to ignore index placement until you plug in actual numbers. If you don't fix them, you'll get spurious factors of $-1$.